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Woody Biomass:Properties and Combustion

FOR 310H S

Bioenergy from Sustainable ForestManagement;Tutorial # 1

28 January 2016; Nicolas Tanguy

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OUTLINE

1) Formation of biofuel (Energy storage)

2) Deformation of biofuel (Energy release)

3) Heating values (HHV, LHV, LHV with moisture)

4) Examples relevant to PS # 1 (Mass conversion and

energy unit conversion)

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FORMATION OF BIOFUEL (1)

•Transformation of Solar Energy to Chemical Energy

• Photosynthesis

• Energy Efficiency

E=(hc)/λ

• Compare to 1st generation

solar panels

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•Composition of woody biomass; Primarily:

(C, H, O)

Cellulose 40-50% (C6H10O5)x

Hemicellulose 20-40% (Xylan,

mannose, galactose, arabinose)

wood

FORMATION OF BIOFUEL (2)

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FORMATION OF BIOFUEL (3)

As well as lipids, protein, minerals

Lignin 20-35% (C31H34O11)n

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ENERGY RELEASE

•Endothermic (absorbs energy)

ENERGY +

•Exothermic (releases energy)

+ ENERGY

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DEFORMATION OF BIOFUEL(BIOMASS BREAKDOWN)

•Metabolism (Catabolism)

- Food chain ( eg. MPB, SBW, decomposition (slow))

•Combustion

- Natural (e.g. Forest fire (rapid))

-Anthropogenic (e.g. Controlled conditions in boiler)

IN ALL CASES, BONDS ARE BROKEN and FORMED

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CALCULATING ENERGY CHANGE ONCOMBUSTION (1)

•∆ =  −  ()

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•∆ =  −  ()

 + 2 →  + 2

∆ = 2628 − 3438 = −810 /

CALCULATING ENERGY CHANGE ONCOMBUSTION (2)

Energy to break down

bonds is 2,628 kJ/mol

Energy to form new

bonds is 3,438 kJ/mol

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COMBUSTION EQUATIONS

Complete combustion (ideal)

Incomplete combustion (real)

OR:

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HEATING VALUES (1)

•Heating value

The amount of energy that

can be released through

combustion

•Wood ~ 20 MJ/kg

•Natural gas ~ 50

MJ/kg

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HEATING VALUES (2) CALORIMETRICVS EFFECTIVE

Calorimetric (higher heating

value) HHV

-Total amount of heat released from

fuel

-Independent of moisture content

Some of the released energy isconsumed during vaporization of H2O

Sources of H2O in wood:

1. Moisture present in the biomass

2. H2O generated when H and Ocombine

Effective (lower heating value)

LHV

- Total amount of heat released minus

the energy used for vaporization

- Affected by presence of moisture

and H in the fuel

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HEATING VALUES (3)

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HOW TO MEASURE THE HHV ?

• Adiabatic Oxygen Bomb Calorimeter (AOB)

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HEATING VALUE OF WOOD (1)

As moisture content increases, the HHV of wood

decreases

Source: Richardson et al. 2002

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HEATING VALUE OF WOOD (2)

LHV = HHV – 2.45 x 0.09 x H

Where:

LHV = effective heating value of dry biomass (MJ/kg dry mass)

HHV = calorimetric heating value of dry biomass (MJ/kg dry mass)

2.45 = amount of heat energy required to vaporize H2O at 20°C (MJ/kg)

0.09 = factor (1 part H + 8 parts O combine to 9 parts H2O)

H = hydrogen content of dry biomass (%)

Example: If hydrogen content of dry biomass is 6 %

LHV is 1.32 MJ/kg less than the HHV

LHV consider the formation of vapor due to recombination of H and O in wood during

combustion

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HEATING VALUE OF WOOD (3)

1.32 MJ/kg

Source: Richardson et al. 2002

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HEATING VALUE OF WOOD (4)

Wem = LHV  – 2.45 x (MC / (100  – MC))

Where:

Wem

= effective heating value of biomass with moisture (MJ/kg dry mass)

LHV = effective heating value of dry biomass (MJ/kg dry mass)

MC = moisture content of biomass on a fresh mass basis (%)

2.45 = amount of heat energy required to vaporize H2O at 20°C (MJ/kg)

Example: At 60 % MC: Wem = Wea – 

3.68

In freshly-felled wood w/a typical MC of 45-58 %, the effective heating value per kg of

dry mass is 15 –20 % lower than the calorimetric value

Wem consider both: the formation of vapor due to recombination of H and O in dry wood

And the energy lost in vaporizing the moisture in wood

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HEATING VALUE OF WOOD (5)

Wem = Wea – 2.45 x (MC / (100 – MC))

3.68 MJ/kg

Source: Richardson et al. 2002

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HEATING VALUE OF WOOD (6)

Source: Richardson et al. 2002

•HHV dry mass basis = ∆

•HHV fresh mass basis = HHV dry mass basis *(1-MCwet)

•LHV dry mass basis = HHV dry mass basis -2.45 x 0.09 x H

•LHV fresh mass basis = LHV dry mass basis *(1-MCwet)

•Wem dry mass basis = LHV  – 2.45 x (MCwet / (100  – MCwet))

•Wem fresh mass basis = LHV fresh mass basis - 2.45 x MCwet

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VARIATIONS IN BIOMASS

•Not all biomass is the same

•Elements distributed differently

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HEATING VALUE OF WOOD (7)

Differences in chemical composition between tree species

= differences in heating values

Lignin, resin and waxes > cellulose and hemicellulose

Softwoods > hardwoodsBranches, bark, foliage > stemwood

Photos by: Daniel Tigner, Canadian Forest Tree Essences

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EMISSIONS

•Low specific emission of CO2• Ratio of CO2 released to amount of

energy

•Depends on C:H• High ratio = high specific emissions

•Natural gas (CH4 gives 0.25)

best of major fuels

• Wood (8.33) is higher but belongs to a

closed system

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Calculating mass of biomass using moisture content

Method 1:

mbs1 x (1 - MCwet1) = mbs2 x (1 - MCwet2)

Where:

mbs1 = mass of biomass sample 1 (in tonnes)

MCwet1 = moisture content of biomass sample 1 on a wet basismbs2 = mass of biomass sample 2 (in tonnes)

MCwet2 = moisture content of biomass sample 2 on a wet basis

Note: moisture content is in decimal form (i.e. 30% is 0.3)

*Use this equation when given any type of question dealing with convertingbiomass from one type to another when given the moisture contents on awet basis (i.e. wood chips to pellets)

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Example

Convert 30 tonnes of wood chips to a mass equivalent in pellets. Moisture

content of wood chips is 45 % and of pellets it is 10 % (wet basis)

Solution:

mbs1 x (1-MCwet1) = mbs2 x (1-MCwet2)

30 x (1 - 0.45) = mbs2 x (1 – 0.10)

mbs2 = [30 x (1-0.45)] / (1 – 0.10)

mbs2 = 18.3 tonnes

Answer: Mass of the pellets is 18.3 tonnes

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Calculating mass of biomass using moisture content

Method 2:

Alternatively, one can calculate mass of samples using the following 2

equations:

A: MCwet = (massH2O / masswet)

-where masswet is the mass of biomass as received

-where massH2O is the mass of the water in the biomass

B: mod = mbs x (1 – MCwet)

-where mod is the mass of the biomass in oven dry tonnes

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Example

Convert 30 tonnes of wood chips to a mass equivalent in pellets. Moisture

content of wood chips is 45 % and of pellets it is 10 % (wet basis).

Solution:

a) Find the mass of water (massH2O) using equation A: MCwet = (massH2O / masswet)

b) Find the oven dry mass of the sample

c) Using equation B we can determine the mass of sample 2 (pellets)

B: mod = mbs x (1 – MCwet)

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A) MCwet = (massH2O / masswet)

- masswet = 30 t (mass of wood chips as received)

- MCwet = 45 % (moisture content of wood chips)

a) Therefore: massH2O = (MCwet x masswet)

massH2O

= (0.45 x 30)

massH2O = 13.5 t

13.5 t of water in the 30 t of biomass

b) Thus the oven dry mass: 30 t – 13.5 t = 16.5 t

16.5 t of oven dry biomass at moisture content 0%

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c) Now we need to convert the oven dry tonnes of biomass to mass of pellets

using the 10 % moisture content of pellets (MCwet) and equation B

B) mod = mbs x (1 – MCwet)

16.5 t = mbs x (1 – 0.1)

mbs = 16.5 t / (1 – 0.1)

mbs = 18.3 t

Answer: Mass of the pellets is 18.3 tonnes

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Unit Conversions

Unit of energy :Joule (kJ, MJ, GJ…), Calories (cal), British thermal unit (Btu),

barrel oil equivalent (Boe)…etc.

Example: Convert 15 GJ to boe

1 Boe = 6.1 GJ

15GJ/6.1 = 2.46 Boe

“https://bioenergy.ornl.gov/papers/misc/energy_conv.html”

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PROBLEM SET #1

• Due date: 4 February 2016 (start of class)

 – Next Thursday

• Nicolas Tanguy’s office hours:

 – Wednesday 3-5 pm (room 2016 ESC)

 – or by appointment

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